# Defining an Inner Product

How can I show that these two formulas for $\langle v,w \rangle$ might define inner products on $\Bbb R^2$?

1.) $v_1 w_2 + v_2 w_1$

2.) $2v_1 w_1 + (v_1-v_2)(w_1-w_2)$

I know that for number 1 it does not define an inner product and for number 2 it does define an inner product but why? I also know in order to define an inner product it must satisfy the conditions of bilinearity, symmetry, and positivity but I am confused on how I can show that?

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what do you mean by v1w2 is it dot product? –  Ram Sep 23 '12 at 7:00
I am sorry but I have no idea. I dont think they are asking for the dot product but to define that these formulas satsify the axioms of an inner product. –  Q.matin Sep 23 '12 at 7:04
I mean $v_1w_2+v_2w_1$ mean $v_1.w_2+v_2.w_1$ –  Ram Sep 23 '12 at 7:09

1) let $(v,w) \equiv v_1 w_2 + v_2 w_1$, then let check bilinearity $(\alpha v, w) = \alpha v_1 w_2 + \alpha v_2 w_1 = \alpha (v_1 w_2 + v_2 w_1) = \alpha (v,w)$ $(v, \beta w) = v_1 \beta w_2 + v_2 \beta w_1 = \beta (v_1 w_2 + v_2 w_1) = \beta (v,w)$ $(v+q,w) = v_1 w_2 + q_1 w_2 + w_1 v_2 + w_1 q_2 = (v,w) + (q,w)$ and so on

2) symmetry $(v,w) = v_1 w_2 + v_2 w_1 = w_1 v_2 + w_2 v_1 = (w,v)$

3) positivity $(v,v) = v_1 v_2 + v_2 v_1 = 2 v_1 v_2$ which is not always $>0$. your definition failed here.

in contrast:

1) let $(v,w) \equiv 2 v_1 w_1 + (v_1 -v_2)(w_1-w_2)$, positivity $(v,v) = 2 v_1 v_1 + (v_1 - v_2)(v_1-v_2) = 2 v_1^2 + (v_1-v_2)^2 > 0$ for any $v \ne 0$

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I understood everything up to number 3 on positivity. I understood how you got 2v1v2 but why is it not always > 0? –  Q.matin Sep 23 '12 at 7:15
Let us select element $v=(1,-1)$ and calculate $(v,v) = 2 \cdot 1 \cdot (-1) = -2$. –  0x2207 Sep 23 '12 at 7:17
I understand now, so in order to satisfy positivity you need it to be of squares of sorts? –  Q.matin Sep 23 '12 at 7:20
Every legal inner product in euclidean space can be expressed as $(x,y) = \sum_{ij} g_{ij} x_{i} y_{j}$, where $g_{ij}$ named 'metric tensor' must be positive defined. $(x,y) \equiv x_1 y_1 + 0.5 x_1 y_2 + 0.5 x_2 y_1 + x_2 y_2$ is also ok. $(x,x) = x_1^2 + x_1 x_2 + x_2^2$ we know, that this is always $>0$ because the square equation has no real roots. –  0x2207 Sep 23 '12 at 7:24
I do not know what a metric tensor is but nonetheless thank you very much!!! –  Q.matin Sep 23 '12 at 7:29
1. What is $< x,x>$ where $x=(1,-1)\in \mathbb{R}^2$?
2. i). $<x,x>=2x_1^2+(x_1-x_2)^2\ge 0\quad\forall x=(x_1,x_2)\in\mathbb{R}^2$. Also note that $<x,x>=0 \Leftrightarrow x=0$.

ii)$<v,w>=<w,v>$, as multiplication of real numbers is commutative.

iii)$<u,v+aw>=2u_1(v_1+aw_1)+(u_1-u_2)(v_1+aw_1-v_2-aw_2)=2u_1v_1+(u_1-u_2)(v_1-v_2)+a[2u_1w_1+(u_1-u_2)(w_1-w_2)=<u,v>+a<u,w>$

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Yup, I understand now thank you very much! –  Q.matin Sep 23 '12 at 7:29